Rotation Of Complex Numbers

Rotation of complex numbers means a geometric transformation that changes its arguments (angle) while preserving its modulus or magnitude. It means that we change the number by multiplying with iota or -1 to change or rotate it . This could also be in degree measures like 180 or 270 or any other measure.. This operation is useful in many fields like engineering, physics, computer graphics, etc. The problem of rotation of vectors is generally solved in the Euler or Polar form of complex numbers.

In this article, we will cover the concept of the rotation of complex numbers. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Vector Notation

Let us take any complex number $\mathrm{z}={x}+\mathrm{i} y$, so point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ represents it on the Argand Plane. Then OP can be represented as a vector $\overrightarrow{O P}=x \hat{i}+y \hat{j}$, where $\hat{i}$ represents the x -axis while $\hat{j}$ represents the y -axis and O is the origin.

Therefore, the complex number z can be represented as $\overrightarrow{O P}$
Similarly, a vector starting from point $A\left(z_1\right)$ and ending at $B\left(z_2\right)$ is represented by $A B$ vector which equals $\left(z_2-z_1\right)$
The length of $A B$ is given by the modulus of this vector $\left|z_2-z_1\right|$

Rotation Theorem (Coni Method)

We know the angle $\theta$. Our purpose is to write down an expression that relates all the four quantities $z_1, z_2, z_3$, and $\theta$.

Consider the vector $z_3-z_2$. Let its argument be $\theta_1$. Similarly, let the argument of the vector $z_1-z_2$ be $\theta_2$. Now, a little thought will show us that $\theta$ is simply:

$
\theta=\theta_1-\theta_2
$

Now, we write $z_3-z_2$ and $z_1-z_2$ in Euler's form:

$
\begin{aligned}
& z_3-z_2=\left|z_3-z_2\right| e^{i \theta_1} \\
& z_1-z_2=\left|z_1-z_2\right| e^{i \theta_2}
\end{aligned}
$

Since we know $\theta_1-\theta_2$, we divide equation (1) by equation (2) to get:

$
\frac{z_3-z_2}{z_1-z_2}=\frac{\left|z_3-z_2\right|}{\left|z_1-z_2\right|} e^{i \theta}
$

This is the relation we were looking for. It relates all the four terms $z_1, z_2, z_3$ and $\theta$.

We are given the vector $z_1-z_2$. We need to modify it into the vector $z_3-z_2$.
Obviously, there will be a change in modulus. Apart from that, we need to rotate the vector $z_1-z_2$ anticlockwise by angle $\theta$ too. This is where the term rotation comes from. Viewing the process in this way, we obtain relation (3) as follows:
1. Write down the unit vector in the direction of the original vector, the one that we need to rotate:

$
\frac{z_1-z_2}{\left|z_1-z_2\right|}
$

2. To rotate this unit vector by an angle $\theta$ (anticlockwise; for clockwise, it will be $-\theta$ ), we multiply it by $e^{i \theta}$.
For the current case, this turns the unit vector into a new unit vector along the direction of the vector $z_3-z_2$ :

$
\frac{z_1-z_2}{\left|z_1-z_2\right|} e^{i \theta}
$

3. Finally, to turn this unit vector into the final vector that we need to obtain after rotation, we multiply the unit vector by the appropriate magnitude.
Thus, we obtain the required final vector after rotation:

$
z_3-z_2=\left|z_3-z_2\right| \cdot \frac{z_1-z_2}{\left|z_1-z_2\right|} e^{i \theta}
$

This is the same relation that we obtained in equation (3).

Using this same approach, suppose that we now wish to relate four complex numbers in the following configuration

Using rotation, we can obtain the vector $z_1-z_2$ from the vector $z_4-z_3$ :
- Unit vector along $z_4-z_3$ direction:

$
\frac{z_4-z_3}{\left|z_4-z_3\right|}
$

(Initial unit vector)
- Unit vector along $z_1-z_2$ direction:

$
\frac{z_1-z_2}{\left|z_1-z_2\right|}
$

(Final unit vector)
- Angle between the two: $\theta$

Thus, we get the final relation:

$
\frac{z_1-z_2}{\left|z_1-z_2\right|}=\frac{z_4-z_3}{\left|z_4-z_3\right|} e^{i \theta}
$

This is the relation we wished to obtain.

Note: The final vector should be in the numerator and the starting vector in the denominator. $\theta$ is positive if rotation is anti-clockwise and negative if it is clockwise.

Summary

The rotation of complex numbers is an important aspect of complex numbers. It helps in shifting the same complex numbers from one point to another. The applications of rotation of complex numbers are also a plus point that helps in many fields of engineering. It is also used variedly in the domains of finance, business, quantum mechanics and computing along with modelling..

Solved Examples Based On the Rotation of Complex Numbers

Example 1: The line joining the origin and the point represented by $z=1+i$ is rotated through an angle $\frac{3 \pi}{2}$ in an anticlockwise direction about the origin and stretched by additional $\sqrt{2}$ unit he point is represented by the complex number

Solution:
As we learned in
Rotation -

$
\frac{z_3-z_1}{z_2-z_1}=\frac{\left|z_3-z_1\right|}{\left|z_2-z_1\right|} \cdot e^{i \Theta}
$

wherein

$
e^{i \theta}=\cos \theta+i \sin \theta
$

using rotation at 0

$
\frac{Z_2-0}{Z_1-0}=\frac{2 \sqrt{2}}{\sqrt{2}} e^{i 3 \frac{\pi}{2}}
$

$
\Rightarrow \frac{Z_2}{Z_1}=2(-i) \Rightarrow Z_2=-2 i(1+i)=2-2 i
$

Hence, the answer is 2-2i.
Example 2: Let $z_1$ and $z_2$ be two roots of the equation $z^2+a z+b=0, z$ being complex further, assume that the origin, $z_1$ and $z_2$ form an equilateral triangle then
1) $a^2=2 b$
2) $a^2=3 b$
3) $a^2=4 b$
4) $a^2=b$

Solution:
As we have learned
Rotation -

$
\frac{z_3-z_1}{z_2-z_1}=\frac{\left|z_3-z_1\right|}{\left|z_2-z_1\right|} \cdot e^{i \Theta}
$

- wherein

$
e^{i \theta}=\cos \theta+i \sin \theta
$
$
\begin{aligned}
& z_1+z_2=-a \\
& z_1 z_2=b
\end{aligned}
$

we have,

$
\begin{aligned}
& \frac{z_1}{z_2}=\frac{\left|z_1\right|}{\left|z_2\right|} e^{i \pi / 3} \\
& =\frac{1}{1} e^{i \pi / 3}
\end{aligned}
$

$
\begin{aligned}
& z_1=z_2 e^{i \pi / 3} \\
& \Rightarrow z_2\left(e^{i \pi / 3}+1\right)=-a \\
& \text { and } \Rightarrow z_2^2\left(e^{i \pi / 3}\right)=b \\
& \therefore b e^{-i \pi / 3}=\frac{a^2}{\left(1+e^{i \pi / 3}\right)^2}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow a^2=b\left(1+e^{i 2 \pi / 3}+2 e^{i \pi / 3}\right) e^{-i \pi / 3} \\
& =b\left(e^{-i \pi / 3}+e^{i(2 \pi-\pi) / 3}+2\right)=b\left(e^{-i \pi / 3}+e^{i \pi / 3}+2\right)=b(2 \cos \pi / 3+2)=3 b
\end{aligned}
$

Hence, the answer is the option 2.

Example 3: Let $z_1, z_2$ be the roots of the equation $z^2+a z+12=0$ and $z_1, z_2$ form an equilateral triangle with the origin. Then, the value of |a| is $\qquad$
Solution:
$0, z_1, z_2$ are the vertex of the equilateral triangle
for equilateral triangle

$
\begin{aligned}
& z_1^2+z_2^2+O^2=z_1 z_2+0+0 \\
& \left(\mathrm{z}_1+\mathrm{z}_2\right)^2=3 \mathrm{z}_1 \mathrm{z}_2 \\
& \Rightarrow\left(-a^2\right)=3(12) \\
& \Rightarrow \mathrm{a}^2=36 \\
& \mathrm{a}=-6 \text { or } 6 \\
& |\mathrm{a}|=6
\end{aligned}
$

Hence, the answer is 6 .
Example 4: The area of the triangle with vertices $A(z), B(i z)$ and $C(z+i z)$ is :
Solution:

$
\begin{aligned}
& A=\frac{1}{2}|z||i z| \\
& A=\frac{|z|^2}{2}
\end{aligned}
$

Hence, the answer is $\frac{1}{2}|z|^2$
Example 5: The line joining the origin and the point represented by $\mathrm{z}=1+\mathrm{i}$ is rotated through an angle $\frac{\pi}{2}$ in an anticlockwise direction about the origin and stretched by additional $\sqrt{3}$ units. The new position of the point is:
1) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(-1+\mathrm{i})$
2) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(1-\mathrm{i})$
3) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(-1-\mathrm{i})$
4) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(1+\mathrm{i})$

Solution
As we have learned in the Rotation Theorem (Coni Method)

Now,

Now,
Let initial point be $\mathrm{A}(1+\mathrm{i})$, O be origin $(0+0 \mathrm{i})$, and $\mathrm{B}(\mathrm{z})$ be the final point

$
|A O|=\sqrt{2}
$

And length of OB is $\sqrt{3}$ more than $|\mathrm{AO}|$, so $|\mathrm{BO}|=\sqrt{2}+\sqrt{3}$
Now using the Rotation Theorem

$
\begin{aligned}
& \frac{O B}{O A}=\frac{|O B|}{|O A|} \cdot e^{\frac{i \pi}{2}} \\
& \frac{z-0}{1+i-0}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} e^{\frac{i \pi}{2}} \\
& z=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} \cdot(1+i) \cdot e^{\frac{i \pi}{2}} \\
& z=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} \cdot(1+i) \cdot(\cos (\pi / 2)+i \sin (\pi / 2)) \\
& z=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} \cdot(1+i) \cdot i \\
& z=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} \cdot(-1+i)
\end{aligned}
$

Hence, the answer is the option 1